Fourier transform

  • Thread starter Benny
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  • #1
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Hi, I'm using the following definition for the Fourier transform.

[tex]
F\left( q \right) = \int\limits_{ - \infty }^\infty {e^{iqx} f\left( x \right)dx}
[/tex]

(I used a capital F instead of f with a squiggle on top because the tex code doesn't seem to be working the way I intended it to.)

I have the function

[tex]
f\left( x \right) = \left\{ {\begin{array}{*{20}c}
{1,a < x < b} \\
{0,otherwise} \\
\end{array}} \right.
[/tex]

So [tex]F\left( q \right) = \int\limits_{ - \infty }^\infty {e^{iqx} f\left( x \right)dx} [/tex]

[tex] = \int\limits_a^b {e^{iqx} dx} [/tex]

[tex]
= \frac{i}{q}\left( {e^{iqa} - e^{iqb} } \right)
[/tex]

According to the definition I'm using, is this the correct answer? I ask this because I'm not given an answer and I need to verify my answer by using the inverse Fourier transform. I haven't done complex analysis so integrals of ratios of exponentials and polynomials aren't things I can deal with right now. Which is why I'd like to know if I've taken the correct approach so that I can at least get through some questions.

Any help would be good thanks.
 
Last edited:

Answers and Replies

  • #2
StatusX
Homework Helper
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Yes, that's correct. You can rewrite that if you want as something like [itex]e^{iq(b+a)/2}\sin(q(b-a))/q[/itex] (that's probably not exactly right). Remember that the fourier transform of unit length square pulse centered at t=0 is a sinc function (sin(x)/x), with some extra normalization factors. Then you just need to shift and broaden this pulse to get a square pulse extending from t=a to t=b. Do you remember the frequency domain operations corresponding to shifting and broadening a pulse in time?
 
  • #3
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I've only just started on Fourier transforms so I don't know too much about them.Thanks for your help though, at least I can get through some questions now.
 

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